Derivation for Moment of Inertia of Rectangle rotated through center

AI Thread Summary
The discussion focuses on deriving the moment of inertia for a rectangle rotated about its center, with the known equation being (1/12)M(a²+b²). The user initially struggles with the integration process and the limits for the double integral. After clarification, they learn that the limits for the inner integral (dx) should range from -a/2 to a/2, and for the outer integral (dy) from -b/2 to b/2. This understanding allows them to successfully complete the derivation. The conversation highlights the importance of mastering double integrals in solving such problems.
matthew487
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Homework Statement



I need to know how to derive the equation for the moment of inertia of a rectangle rotated about an axis through its center. The rectangle has sides a and b. I know the equation to be (1/12)M(a2+b2), but I am having trouble deriving it. I have searched all over the internet without finding any helpful solutions.

Homework Equations



dI = ∫r2dm
dm = λdA

The Attempt at a Solution



I started by taking a point mass of the rectangle to be the distance r from the center. This point would have the mass dm.

dm would then equal λdA
λ would equal M/A or M/ab
dA would be the area of the small point. I'm guessing this would be dydx

So, for dm we have (M/ab)(dydx)

Now for r2. By pythagorean thereom, r2 = x2+y2.

Substituting these into the equation ∫r2dm, we have (M/ab)∫(x2+y2)dydx

I'm pretty sure this is incorrect, but even if it is correct I have no idea how to integrate it. I don't even know what limits to integrate about.
 
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matthew487 said:

Homework Statement



I need to know how to derive the equation for the moment of inertia of a rectangle rotated about an axis through its center. The rectangle has sides a and b. I know the equation to be (1/12)M(a2+b2), but I am having trouble deriving it. I have searched all over the internet without finding any helpful solutions.

Homework Equations



dI = ∫r2dm
dm = λdA

The Attempt at a Solution



I started by taking a point mass of the rectangle to be the distance r from the center. This point would have the mass dm.

dm would then equal λdA
λ would equal M/A or M/ab
dA would be the area of the small point. I'm guessing this would be dydx

So, for dm we have (M/ab)(dydx)

Now for r2. By pythagorean thereom, r2 = x2+y2.

Substituting these into the equation ∫r2dm, we have (M/ab)∫(x2+y2)dydx

I'm pretty sure this is incorrect, but even if it is correct I have no idea how to integrate it. I don't even know what limits to integrate about.

Hi Matthew487...

Welcome to PF!

The item in red should be (M/ab)∫∫(x2+y2)dydx

The limit of the inner integral will be values 'y' will be taking .The limit of outer integral will be values 'x' will be taking.

Considering origin to be at the center .What should be the limits of the inner integral ?What should be the limits of the outer integral ?
 
Ahhhh! Thank You! I had never done or even heard of a double integral before. That was the problem. After quickly learning what that is, I was able to solve it. For the inside dx integral the limits are from -a/2 to a/2. The outside integral for dy is from -b/2 to b/2. Thanks a ton!
 
Excellent! :thumbs:
 
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